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In the LMFDB database, there are 337912 elliptic curves over $\mathbb{Q}$ for which the rank is 1 and the conductor is a prime number.

All of these curves have trivial torsion group.

Is there a known proof to the effect that any such curve must have trivial torsion?

Regardless of whether a proof is known or not - is this phenomenon connected to some more general story?

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    $\begingroup$ Yes. It is known that a curve of prime conductor $p$ and nontrivial torsion either has $p \in \{11, 17, 19, 37\}$ or is a "Setzer-Neumann curve" with $p = n^2 + 64$ and torsion group Z/2Z. It is also known that every such curve has rank zero. So, a curve of prime conductor and positive rank (not just rank 1) must have trivial torsion. $\endgroup$ Commented Jul 10, 2022 at 0:04
  • $\begingroup$ Great, thanks a lot :-) $\endgroup$ Commented Jul 10, 2022 at 0:54
  • $\begingroup$ I'll venture a followup here instead of asking a new question. What I really need is to know that there is an infinite number of rank 1 curves with prime conductor. This seems exceedingly plausible, but I have no idea how to prove such a statement. Do you? $\endgroup$ Commented Jul 10, 2022 at 3:44
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    $\begingroup$ That's surely true but might be beyond the current state of the art. The problem is that it might be already too hard to prove that there are infinitely many elliptic curves of prime conductor without any further constraint! (If $p = n^2 + 64$ then there's an isogenous pair of Setzer-Neumann curves of conductor $p$, but -- aside of the fact that these curves happen to have rank zero, while you want rank 1 -- we can't yet prove that there are infinitely many primes of that form.) $\endgroup$ Commented Jul 10, 2022 at 17:42
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    $\begingroup$ There's at least heuristics and extensive data about the number of elliptic curves of prime conductors. E.g., see Bennett et al's "Computing elliptic curves over Q" and BMSW's "Average ranks of elliptic curves" Bulletin article. $\endgroup$
    – Kimball
    Commented Jul 11, 2022 at 7:51

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